The equation of a circle $C$ is $x^2+y^2-14y+40 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2) + (y^2-14y) = -40$ $(x^2) + (y^2-14y+49) = -40 + 0 + 49$ $x^2 + (y-7)^{2} = 9 = 3^2$ Thus, $(h, k) = (0, 7)$ and $r = 3$.